Let $V$ be a vector space of functions $[0,1]\rightarrow \Bbb R$. What is the maximal possible dimension of $U\subseteq V$, a subspace consisting of monotone functions in $V$?
I was thinking to approach this question using elementary set theory which I learned to find the cardinality of $U$, but that wouldn't be very possible on the interval $[0,1]$.
I've never seen a question like this so I don't know what else can be done.
On
The cardinality of the monotone real functions defined on $[0,1]$ is equal to $\mathfrak{c}$. See What is the cardinality of a set of all monotonic functions on a segment [0,1]?
Hence the dimension of $U$ is at most $\mathfrak{c}$. Now you'll be able to prove that the family of functions
$$f_\alpha(x)= \begin{cases} 0 & 0 \le x <\alpha\\ 1 & \alpha \le x \le 1 \end{cases}$$ for $\alpha \in (0,1)$ is linearly independent (consider the discontinuities of a linear combination). The cardinality of $(f_\alpha)_{\alpha \in (0,1)}$ is $\mathfrak c$. Hence, the dimension of $U$ (over $\mathbb R$) is equal to $\mathfrak{c}$ (while the dimension of $V$ is equal to $2^{\mathfrak c}$).
Hint: If $f$ and $g$ are monotonic functions and none of them is a multiple of the other one, then there are real numbers $\alpha$ and $\beta$ such that $\alpha f+\beta g$ is not monotonic.